Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 20 / a

Let $a_n=2^n/n!$. The ratio test gives $a_{n+1}/a_n=2/(n+1)\to 0<1$, so the series converges absolutely. In fact the exponential series identifies its value aswith the $n=0$ term subtracted.

Here $0<\ln 2<1$, so $(\ln 2{)}^{-n}$ grows exponentially rather than decaying. The positive terms $a_n=n^3/(\ln 2{)}^n$ do not tend to zero; indeed they tend to $+\infty$. Therefore the series diverges by the necessary condition that the terms of a convergent series tend to zero.

The ratio test gives the same conclusion: $a_{n+1}/a_n=((n+1)/n{)}^3/\ln 2\to 1/\ln 2>1$.